Energy absorbed by solution. Q=mc(theta)

[coconut62]

Homework Statement

Please refer to the image.

I know how to get the answer, which is 40 x 4.18 x 1.5 = 250.8 J, but I don't understand why the volume of the 8g of sodium chloride is not taken in as part of the whole volume? Shouldn't the total volume be 40+8 = 48?

Homework Equations

Q=mcθ

 

[BruceW]

8g is not volume. but you are right in that the new volume of the mixture would be very slightly greater than the volume of the water alone. I guess you are meant to assume it is negligible. You could look up the density of sodium chloride, and try to work out how much the volume would change by.

 

[Borek]

Final volume would be 42.8 mL.

Which doesn't help much, as it doesn't tell anything about the solution heat capacity.

 

[BruceW]

ah, that's a bigger change in volume than I thought it would be... I am still guessing that (approximately) constant volume is meant to be assumed ? Also, I am guessing that the constant volume heat capacity is given somewhere else in the question, which is why the OP has written down the equation with the constant volume heat capacity in it. But that's a good point, maybe a clue is in exactly what constants the question gives..

 

[coconut62]

The question gives the specific heat capacity of the solution as 4.18.

 

[Borek]

4.18 J g^-1 K^-1 is a specific heat of a pure water. It is OK to use this value for diluted solutions, as long as you are aware of the fact it is only approximate. For me 16.6% doesn't qualify as diluted.

 

[coconut62]

So my book is actually misleading. In my class test a few days ago a question wanted us to calculated the heat released when 1g of calcium dissolves in 200g of water. I used 200g as the total mass, and my friends all used 201.

Sad.

 

[coconut62]

4.18 J g^-1 K^-1 is a specific heat of a pure water. It is OK to use this value for diluted solutions, as long as you are aware of the fact it is only approximate. For me 16.6% doesn't qualify as diluted.

Wait, how did you get that 16.6%?

And the 42.8mL?

 

[Borek]

8g and 40 mL of water means 8 g in 48 g, which is 8/48*100%=16.6%.

42.8 mL is slightly more complicated, as it requires checking the solution density in tables. Done using my concentration calculator, see attached screenshot. I have entered mass of the solute (8 g) and total solution mass (Sm = 48). Program checks the density of the 16.66% solution in tables, then uses this density to calculate volume (42.8024 mL).

 

[coconut62]

Ah, I see. Thanks.

 

[BruceW]

hmm. I was thinking about it, and effectively the book's answer gives the amount of heat required to just change the temperature of the water. Therefore, the volume of the water and the specific heat capacity of the water is used. Now this is a good approximation, because the specific heat capacity of the sodium chloride is roughly 5 times less than that of the water. So even though the volume of sodium chloride is significant, its low specific heat capacity (relative to water) means that the heat required to change the temperature of the sodium chloride is much less than the heat required to change the temperature of the water. So they have used the approximation that the total heat is approximately the same as the heat required for just the water.

Now, this doesn't take into consideration that the solution is going to be physically different than just a mixture of the water and sodium chloride. (and in reality, the solution is going to be different). So the argument isn't really water-tight. But I think maybe the last paragraph is the logic that the book has used... But its difficult, and its not easy to see exactly what they want for an answer.